| Book Name | : | Target Book |
| Paper Type | : | Model Test Question Paper (2024 – 2025) |
| Paper Category | : | Madhyamik Suggestion |
Question – 1
Choose the correct answer for each of the following questions:
(i) The simple interest at x % for x years will be ₹ x for a principal amount of
(a) ₹ x
(b) ₹ 100x
(c) \frac{100}{x}
(d) ₹ \frac{100}{x^2}
Answer
(c) P = 100\over x
Explanation
Simple Interest (I) = \frac{\text{P × r × t}}{100}
or, x = \frac{\text{P × x × x}}{100}
or, P = \frac{\text{ x × 100}}{x × x}
or, P = 100\over x
(ii) If x² – 3x + 1 = 0, then the value of x⁵ + 1\over \text{x⁵} is—
(a) 87
(b) 123
(c) 135
(d) 201
Answer
(b) 123
(iii) AB and CD are two parallel chords of length 16 cm each. If the radius of the circle is 10 cm then find the distance between the chords.
(a) 10 cm
(b) 12 cm
(c) 16 cm
(d) None of these
Answer
(b) 12 cm
Explanation
d1 = \sqrt{\text{r}^2 - ({\text{l}\over2})^2}
= \sqrt{\text{10}^2 - ({\text{16}\over2})^2}
= \sqrt{\text{100} - 64} = 6 cm
d2 = \sqrt{\text{r}^2 - ({\text{l}\over2})^2}
= \sqrt{\text{10}^2 - ({\text{16}\over2})^2}
= \sqrt{\text{100} - 64} = 6 cm
dtotal = d1 + d2 = 6 + 6 = 12 cm
(iv) The value of 4π\over 15 radian is—
(a) 24°
(b) 48°
(c) 36°
(d) 72°
Answer
(b) 48°
Explanation
4π\over 15 = 4 × 180º\over 15 = 48°
(v) If the ratio of the heights of two solid right circular cylinders of equal volume is 1 : 2 then the ratio of the lengths of the radii of the two cylinders will be:
(a) 1 : √2
(b) √2 : 1
(c) 1 : 2
(d) 2 : 1
Answer
(b) √2 : 1
Explanation
Since the two cylinders have equal volume, we can write:
πr12h1 = πr22h2
or,
or, r1 : r2 = √2 : 1
(vi) If the classes of a frequency distribution are 1–5, 6–10, 11–15, 16–20, then the length of each class will be —
(a) 4
(b) 4.5
(c) 5
(d) 2
Answer
(c) 5
Explanation
Given
Class Interval: 1–5, 6–10, 11–15, 16–20
Class Limit: 0.5 – 5.5, 5.5 – 10.5, 10.5 – 15.5, 15.5 – 20.5
Class length = 5.5 – 0.5 = 5
Question – 2
Fill in the blanks (Any Five):
(i) Current population of a village is p. If the rate of growth of population is 2% per year then the population of the village after 2 years will be ____.
(ii) If a : 2 = b : 5 = c : 8 then 50% of a = 20% of b = ____% of c.
(iii) The maximum number of tangents can be drawn to a circle from an external point is ____.
(iv) If 2 cos 3θ = 1 then sin 3θ\over 2 = _____.
(v) If the length of the diameter and height of a right circular cylinder are 3 cm and 4 cm respectively, then the length of the longest rod that can be placed inside that cylinder is ____.
(vi) Mode of the data 2, 3, 9, 10, 3, 9, 9 is ____.
Answer
(i) 1.0404p
(ii) 12.5 %
(iii) 2
(iv) 1\over 2
Explanation
2 cos 3θ = 1
or, cos 3θ = 1\over2
or, cos 3θ = cos 60º
or, 3θ = 60º
or, θ = 20º
Now, sin 3θ\over 2 = sin 3 × 20º\over 2
= sin 30º = 1\over 2
(v) 9
Question – 3
State whether True or False (Any Five):
(i) If in a partnership business the ratio of the capitals of Raju and Asif is 5 : 4. If Raju earns ₹100 as his share of profit then Asif earns ₹90 as his share of profit.
(ii) If √a : √b = √(ab), then a and b are real numbers.
(iii) Two similar triangles are always congruent.
(iv) If 0° ≤ α < 90° then the least value of sec²α + cos²α is 2.
(v) If the of the base of a right circular cone is reduced to half and the height is doubled then volume of the cone remains the same.
(vi) Median of the first 10 natural numbers is 5.5.
Answer
(i) False
Explanation
Profit ratio = ₹100 : ₹90 = 10 : 9
Capital ratio = 5 : 4
Profit ratio ≠ Capital ratio
(ii) False
(iii) False
(iv) True
(v) False
(vi) True
Explanation
The first 10 natural numbers are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
For an even number of observations, the median is the average of the two middle values.
Median = \text{5th value + 6th value}\over 2
= 5 + 6\over 2
= 11\over 2
= 5.5
Question – 4
Answer the following Questions (Any Ten):
(i) Ratio of the capitals of three persons is 4 : 7 : 9. If the profit earned by the first person is ₹100 less than that by the third person then find the profit earned by the second person.
Answer
Ratio of the capitals = 4 : 7 : 9
Profit of A = 4x
Profit of C = 9x
ATP: 9x – 4x = 100
or, 5x = 100
or, x = 20
Profit of B = 7x = 7 × 20 = ₹ 140
(ii) At what rate of simple interest percent per annum, the interest on a certain principal for 4 years becomes \frac{8}{25} of the principal.
Answer
time = 4 years
Principal = P
= \frac{8}{25}P
Rate (r) = {8\over25}P × 100\over P × 4 = 8 %</p> </div> <p>(iii) Which one of (√15 + √3) and (√10 + √8) is greater?</p> <p><strong>Answer</strong></p> <div style="background-color: #e0e4fa; border-radius: 3px; padding: 5px; margin-bottom: 10px; color: #010013;"> <p>√15 + √3 = 3.873 + 1.732 = 5.605</p> <p>√10 + √8 = 3.162 + 2.828 = 5.990</p> <p>(√10 + √8) > (√15 + √3)</p> </div> <p>(iv) If α and β are the roots of the equation x² - 22x + 105 = 0, then find the value of (α - β).</p> <p><strong>Answer</strong></p> <div style="background-color: #e0e4fa; border-radius: 3px; padding: 5px; margin-bottom: 10px; color: #010013;"> <p>α + β = 22</p> <p>α × β = 105</p> <p>(α - β)² = (α + β)² - 4(αβ)</p> <p>or, (α - β)² = 22² - 4 × 105</p> <p>or, (α - β)² = 22² - 4 × 105</p> <p>or, (α - β)² = 484 - 420</p> <p>or, α - β = √64 = 8</p> </div> <p>(v) In the adjacent diagram O is the centre of a circle with diameter AB. If ∠AOD = 140° and ∠CAB = 50°, then find the measure of ∠BED.</p> <p><img class="size-medium wp-image-62876 aligncenter" style="background-color: #f0f0f0; padding: 10px; border-radius: 5px;" src="https://flasheducation.online/wp-content/uploads/2024/12/ABCD_is_a_cyclic_quadrilateral._AB_and_DC_are_extended_to_meet_at_P._Prove_that_PA_PB___PC_PD__1_-removebg-preview-e1734284222394-300x195.png" alt="ABCD_is_a_cyclic_quadrilateral._AB_and_DC_are_extended_to_meet_at_P._Prove_that_PA_PB___PC_PD__1_-removebg-preview" width="300" height="195" /></p> <p><strong>Answer</strong></p> <div style="background-color: #e0e4fa; border-radius: 3px; padding: 5px; margin-bottom: 10px; color: #010013;"> <p>Given that ∠AOD = 140°</p> <p>∴ reflex ∠AOD = 360° - 140° = 220°</p> <p>Again, the central angle produced by the arc ÂBD is reflex ∠AOD and angle in circle = ∠ACD</p> <p>∴ reflex ∠AOD = 2∠ACD</p> <p>or, 220° = 2∠ACD</p> <p>∴ ∠ACD = 220°\over2 = 110°
Given that ∠CAB = 50°
Now, in ΔACE,
∠AEC + ∠EAC + ∠ACE = 180°
or, ∠AEC = 180° - 50° - 110° = 20°
Hence ∠BED = 20°
(vi) Two equal circles, each of radius 10 cm intersect each other and their common chord is of length 12 cm. What is the distance between their centres?
Answer
Radius of each circle (r) = 10 cm
Length of the common chord (L) = 12 cm
Let the distance between the centers of the circles be d. The common chord divides the distance dd into two equal parts at its midpoint (perpendicular bisector). The relationship is:
r = \sqrt{({\text{L}\over2})^2 - ({\text{d}\over2})^2}
or, 10 = \sqrt{({\text{12}\over2})^2 - ({\text{d}\over2})^2}
or, ({\text{d}\over2})^2 = 100 - 36 = 64
or, ({\text{d}\over2})^2 = 64
or, ({\text{d}\over2}) = √64 = 8 cm
or, d = 2 × 8 cm = 16 cm
(vii) In ΔABC, AB = 6 cm, BC = 8 cm and AC = 12 cm. D is a point on AC such that ∠ADB = ∠ABC. Then find the length of BD.
Answer
In ΔABD and ΔABC,
∠BAD = ∠BAC (Common)
∠ADB = ∠ABC (Given)
So ΔABD and ΔABC are similar
Hence: \text{BD}\over \text{BC} = \text{AB}\over \text{AC}
or, BD = \text{AB}\over \text{AC} × BC = \text{6}\over \text{12} × 8 = 4 cm
(viii) Find the circular measure of 111\over4º
Answer
111\over4º = 45\over4º
= 45\over4 × π\over180
= π\over16
(ix) Find the value of cos 18° (cos 72° cos² 22° + 1\over \text{tan 72º sec² 68º} )</p> <p><strong>Answer</strong></p> <div style="background-color: #e0e4fa; border-radius: 3px; padding: 5px; margin-bottom: 10px; color: #010013;"> <p>We know,</p> <ul> <li>cot 72° = cot (90° - 18°) = tan 18°</li> <li>tan 72° = tan (90° - 18°) = cot 18°</li> </ul> <p>cot 18° (cot 72° cos² 22° + 1\over \text{tan 72º sec² 68º} )</p> <p>= cot 18° cot 72° cos² 22° + cot 18°1\over \text{tan 72º sec² 68º} </p> <p>= cot 18° tan 18° cos² 22° + cot 18°1\over \text{cot 18° sec² 68º} </p> <p>= cos² 22° + cos² 68º</p> <p>= cos² 22° + cos² (90º - 22º)</p> <p>= cos² 22° + sin² 22º = 1</p> </div> <p>(x) Curved surface area of a solid sphere is S and volume V, then find the value of S^3\over V^2
Answer
Given a solid sphere with radius r, we know:
Surface area S = 4πr²
Volume V = (4/3)πr³
Now, S^3\over V^2.
= (4πr²)^3\over ({({4\over3})}πr³)^2
= 64π³r⁶\over {({16\over9})}π²r⁶
= 64π\over {({16\over9})}
= 64π × 9\over 16
= 36 π
(xi) If each edge of a cube is increased by 50% then what is the percentage increase in its total surface area.
Answer
Let the edge = a cm.
New measure of the edges after increase
= a + 50 %a = 1.5a
Total surface area of the original cube = 6a²
Total surafec area of the new cube
= 6 × (1.5a)²
= 2.25 × 6a²
= 2.25 × Total surafec area of the original cube
Increase in the area
= New total surface area - orignal suface area
= (2.25−1) × 6a² = 1.25 × 6a²
∴ Percentage increase in surface area = 125 %
(xii) If u = \text{x}_i -25\over10, ∑fiui = 20 and ∑fi = 100 then find x̄.
Answer
Given: ui = x_i - 25\over10, ∑fiui = 20 and ∑fi = 100
Now, ui = x_i - A\over10 = x_i - 25\over10
Therefore, h = 10 and A = 25
x̄ = A + h{{1\overN}}∑fiui
x̄ = 25 + 10{{1\over100}} × 20 = 25 + 2 = 27
Question - 5
Answer any one of the following Questions:
(i) Due to an anti-smoking campaign, the rate of yearly decrease in the number of smokers is 6 1\over4 %. If the number of smokers in a town is 33,750, then what was the number of smokers in that town 3 years before?
(ii) Ayan, Abhijit and Aditi start a partnership business with capitals of ₹ 6,000, ₹ 8,000 and ₹ 9,000 respectively. After few months Ayan invests ₹ 3,000 more. If the total profit at the end of the year is ₹ 3,000 and Aditi gets ₹ 1,080 as her share of profit then find the number of months after which Ayan invested ₹ 3,000 more.
Answer
Present Population (V) = ₹ 33,750
rate (r) = 25\over4%
time (n) = 3 years
Initial Population (Vo) = ?
Present Population (V) = Vo(1 - 25\over4×100)3
or, 33750 = Vo (1 - 1\over 16)3
or, 33750 = Vo (15\over 16)3
or, Vo = 33750 × 16 × 16 × 16\over 15 × 15 × 15 = 40,960.
The number of smokers three years ago was approximately 40,960.
Question - 6
Answer any one of the following Questions:
(i) Solve:
\frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6}
Answer
\frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6}
or, \frac{1}{x-2} - \frac{1}{x-1} + \frac{1}{x-3} + \frac{1}{x-2}+\frac{1}{x-4} - \frac{1}{x-3} = \frac{1}{6}
or, \frac{1}{x-4} - \frac{1}{x-1} = \frac{1}{6}
or, \frac{3}{(x-4)(x-1)} = \frac{1}{6}
or, 18 = x² - 5x + 4
or, x² - 5x - 14 = 0
or, x² - 7x + 2x - 14 = 0
or, x(x - 7) + 2(x - 7) = 0
or, (x + 2)(x - 7) = 0
x = 7 or -2
(ii) From the equation the roots of which are the squares of the roots of the equation .
Answer
Let the roots of the equation x² + x + 1 be α and β
α + β = - 1
α×β = 1
Now, α² + β² = (α + β)² - 2α×β
= (- 1)² - 2(1)
= - 1
and α²×β² = (αβ)² = 1
New equation:
x² - (α² + β²)x + (αβ)² = 1
or, x² + x + 1 = 0
Question - 7
Answer any one of the following Questions:
(i) If a = \frac{\sqrt{5}+1}{\sqrt{5}-1} and b = \frac{\sqrt{5}-1}{\sqrt{5}+1} then evaluate \frac{3 a^{2}+5 a b+3 b^{2}}{3 a^{2}-5 a b+3 b^{2}}
Answer
a + b = \frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^{2}+(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}
= \frac{5+1+2 \sqrt{5}+5+1-2 \sqrt{5}}{5-1}=\frac{12}{4}=3
\ \& \ a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}
ab = \frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1
Now, \frac{3 a^{2}+5 a b+3 b^{2}}{3 a^{2}-5 a b+3 b^{2}}=\frac{3 a^{2}+6 a b+3 b^{2}-a b}{3 a^{2}-6 a b+3 b^{2}+a b}=\frac{3\left(a^{2}+2 a b+b^{2}\right)-1}{3\left(a^{2}-2 a b+b^{2}\right)+1}
= \frac{3(a+b)^{2}-a b}{3(a-b)^{2}+a b}=\frac{3(3)^{2}-1}{3(\sqrt{5})^{2}+1}=\frac{27-1}{3.5+1}=\frac{26}{16}=\frac{13}{8}
= 1 \frac{5}{8}
(ii) If a,b,ca, b, ca,b,c and b,c,ab, c, ab,c,a then show that (a3b3+b3c3+c3b3)=abc(a3+b3+c3)(a^3b^3 + b^3c^3 + c^3b^3) = abc(a^3 + b^3 + c^3)(a3b3+b3c3+c3b3)=abc(a3+b3+c3).
Answer
Question - 8
Answer any one of the following Questions:
(i) If ay−bxby+cz=cx−azcz+ax=bz−cyax+by\frac{ay - bx}{by + cz} = \frac{cx - az}{cz + ax} = \frac{bz - cy}{ax + by}by+czay−bx=cz+axcx−az=ax+bybz−cy, then show that xa=yb=zc\frac{x}{a} = \frac{y}{b} = \frac{z}{c}ax=by=cz.
Answer
(ii) If x2by+cz=y2cz+ax=z2ax+by=1\frac{x^2}{by + cz} = \frac{y^2}{cz + ax} = \frac{z^2}{ax + by} = 1by+czx2=cz+axy2=ax+byz2=1, then show that aa+x+bb+y+cc+z=1\frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1a+xa+b+yb+c+zc=1.
Answer
Question - 9
Answer any one of the following Questions:
(i) Prove that the angle which an arc of a circle subtends at the centre is twice the angle subtended by it at any point in the remaining part of the circle.
Answer
(ii) If a perpendicular is drawn from the right angle to the hypotenuse in a right angled triangle, then prove that the triangles on each side of the perpendicular are similar and also similar to the whole triangle.
Answer
Question - 10
Answer any one of the following Questions:
(i) Chords AC and BD of a circle intersect each other at the point O. If the tangents to the circle at the points A and B intersect each other at P and the tangents to the circle at the points C and D intersect each other at point Q, then prove that ∠APB+∠CQD=2∠BOC\angle APB + \angle CQD = 2\angle BOC∠APB+∠CQD=2∠BOC.
Answer
(ii) In ARST, ∠S\angle S∠S is a right angle if X and Y are the middle points of the sides RS and ST, then prove RY2+XT2=5XY2RY^2 + XT^2 = 5XY^2RY2+XT2=5XY2.
Answer
Question - 11
Answer any one of the following Questions:
(i) Draw a △ABC of sides BC = 7 cm, AB = 5 cm and AC = 6 cm. Hence draw the incircle of the triangle. (Only traces of construction are to be given).
Answer
(ii) Find geometrically the mean proportion of 10 cm and 4 cm. (Only traces of construction are to be given).
Answer
Question - 12
Answer any two of the following Questions:
(i) Find the value of 1+sinθ1−sinθ+1−sinθ1+sinθ\frac{\sqrt{1 + \sin \theta}}{\sqrt{1 - \sin \theta}} + \frac{\sqrt{1 - \sin \theta}}{\sqrt{1 + \sin \theta}}.
Answer
(ii) If cosθ+secθ=2\cos \theta + \sec \theta = 2, then evaluate cos3θ+sec3θ\cos^3 \theta + \sec^3 \theta.
Answer
(iii) Show that tan15∘+tan75∘=sec215∘sec215∘−1\tan 15^\circ + \tan 75^\circ = \frac{\sec^2 15^\circ}{\sqrt{\sec^2 15^\circ - 1}}.
Answer
Question - 13
Answer any one of the following Questions:
(i) From the roof of a house of 11 m high, the angle of depression of the top and the foot of a lamp post are observed 30° and 60° respectively. Find the height of the lamp post.
(ii) From a point on a 5√3 m high railway overbridge Rohit finds that the angle of depression of a train engine on one side of the bridge is 30° and after 2 seconds the engine is found on the other side of the over-bridge, at an angle of depression of 45°. Find the speed of the train.
Question - 14
Answer any two of the following Questions:
(i) Four cubical holes are to be dug out from the four corners of a rectangular field of length 20 m and breadth 15 m. The soil from the holes are evenly spread on the rectangular field as a result the field is raised. Find the increase in height of the entire rectangular field.
Answer
(ii) The external and internal diameters of a hollow right circular cylinder of height 36 cm are 16 cm and 12 cm respectively. Find the number of solid cylinders, each of diameter 2 cm and length 6 cm can be made by melting and recasting the hollow cylinder.
Answer
(iii) Area of the base of a right circular conical tent is 13.86 sq.m. Tarpaulin worth ₹ 5,775 is required to make the tent. If the cost of tarpaulin is ₹ 250 per m², then find the height of the tent and also the volume of the air inside the conical tent.
Answer
Question - 15
Answer any two of the following Questions:
(i) If the mean of the following frequency distribution is 67.45 then find the value of f2:
| Class limit | Frequency |
| 60 - 62 | 15 |
| 63 - 65 | 54 |
| 66 - 68 | f2 |
| 69 - 71 | 81 |
| 72 - 74 | 34 |
Answer
(ii) Find the median from the following frequency distribution table:
| Class limit | Frequency |
| 10 - 19 | 8 |
| 20 - 29 | 11 |
| 30 - 39 | 15 |
| 40 - 49 | 17 |
| 50 - 59 | 17 |
| 60 - 69 | 12 |
Answer
(iii) From the following frequency distribution table, find the mode:
| Class limit | Frequency |
| 56 - 60 | 7 |
| 61 - 65 | 25 |
| 66 - 70 | 43 |
| 71 - 75 | 25 |
| 76 - 80 | 7 |
Answer